Title Description
Is given a length of not more than 200 is 2 0 letter string lowercase English letters 0 (the string in each row 20 is 2 0 letter input mode, and to ensure that each row is constant 20 is 2 0). This requires the letter string is divided into K K parts, and the number of words included in each add up to the total number of the maximum.
Each word may be contained partially overlap. After the election with a word, its first letter can not be reused. For example, the string this may be included this and is, selected this after can not be contained th.
Does not exceed a given word in 6 six words in the dictionary.
The maximum number of the required output.
Input Format
The first row of each group has two positive integers P, K P , K. p rows and p represents string, K K represents a divided K K parts.
Next the p- the p-lines, each line has 20 2 0 characters.
Then there is again a positive integer S S, is the number of dictionary words. The next S S lines, each line has a word.
Output Format
. 1 an integer, respectively corresponding to each set of test data.
Sample input and output
Thinking
Meaning of the questions have a place under the influence of said relatively vague, is the overlap situation.
After this you can actually use the re-election is, but can not choose the first letter of t, which is optional full main_len can not be elected.
After about I got it to know that this is a recursive substring find string in the meet can find the word in the dictionary and can appear a few words.
A sum [i] [j] in the string to a front section i ~ j within a few words.
With F [i] [j] to represent the maximum number of characters in the i-th word segment j cut obtainable.
Obviously it can be transferred as follows:
f [ i ] [ k ] = max ( f [ i ] [ k ] , f [ j ] [ k - 1] + sum [ i - 1 ] [ j ] ) ; 其中 k ∈ [ 2, min ( k, j + 1 ) ] ;
CODE
#include
<
bits/stdc++.h
>
#define
dbg(
x
) cout
<<
#x
<<
"
=
"
<< x
<< endl
#define
eps
1
e
-
8
#define
pi
acos(
-
1.0
)
using
namespace std
;
typedef
long
long LL
;
const
int inf
=
0x
3f3f3f3f
;
template
<
class T
>
inline
void
read(T
&
res
)
{
char c
;T flag
=
1
;
while((c
=
getchar())
<
'
0
'
||c
>
'
9
'
)if(c
==
'
-
'
)flag
=-
1
;res
=c
-
'
0
'
;
while((c
=
getchar())
>=
'
0
'
&&c
<=
'
9
'
)res
=res
*
10
+c
-
'
0
'
;res
*=flag
;
}
namespace _buff
{
const
size_t BUFF
=
1
<<
19
;
char
ibuf
[BUFF
],
*ib
= ibuf
,
*ie
= ibuf
;
char
getc()
{
if
(ib
== ie
)
{
ib
= ibuf
;
ie
= ibuf
+
fread(ibuf
,
1
, BUFF
, stdin
);
}
return ib
== ie
?
-
1
:
*ib
++
;
}
}
int
qread()
{
using
namespace _buff
;
int ret
=
0
;
bool pos
=
true
;
char c
=
getc();
for
(;
(c
<
'
0
'
|| c
>
'
9
'
)
&& c
!=
'
-
'
; c
=
getc())
{
assert(
~c
);
}
if
(c
==
'
-
'
)
{
pos
=
false
;
c
=
getc();
}
for
(; c
>=
'
0
'
&& c
<=
'
9
'
; c
=
getc())
{
ret
=
(ret
<<
3
)
+
(ret
<<
1
)
+
(c
^
48
);
}
return pos
? ret
:
-ret
;
}
int p
, k
;
string a
, b
;
string
ss
[
15
];
int
sum
[
1000
][
1000
];
int
f
[
1000
][
1000
];
int
main()
{
read(p
);read(k
);
for
(
int i
=
1
; i
<= p
;
++i
)
{
cin
>> b
;
a
+= b
;
}
int main_len
=
a
.size();
int n
;
read(n
);
vector
<int>
len(n
+
7
);
for
(
int i
=
1
; i
<= n
;
++i
)
{
cin
>>
ss
[i
];
len
[i
]
=
ss
[i
].size();
}
for
(
int i
=
0
; i
< main_len
;
++i
)
{
for
(
int j
= i
; j
< main_len
;
++j
)
{
//todo:枚举子段
for
(
int k
= i
; k
<= j
;
++k
)
{
int ok
=
1
;
for
(
int z
=
1
; z
<= n
;
++z
)
{
if(k
+
len
[z
]
-
1
> j
)
continue;
for
(
int w
=
0
; w
<
len
[z
];
++w
)
{
if(w
==
len
[z
]
-
1
&&
a
[k
+ w
]
==
ss
[z
][w
])
{
ok
=
0
;
break;
}
if(
a
[k
+ w
]
!=
ss
[z
][w
])
{
break;
}
}
if(
!ok
)
break;
}
if(
!ok
)
{
sum
[i
][j
]
++
;
}
}
}
}
// for ( int i = 0; i < main_len; ++i ) {
// for ( int j = 0; j < main_len; ++j ) {
// printf("sum[%d][%d]:%d\n",i, j, sum[i][j]);
// }
// }
for
(
int i
=
0
; i
< main_len
;
++i
)
{
f
[i
][
1
]
=
sum
[
0
][i
];
for
(
int j
=
0
; j
< i
;
++j
)
{
for
(
int k
=
2
; k
<=
min(k
, j
+
1
);
++k
)
{
f
[i
][k
]
=
max(
f
[i
][k
],
f
[j
][k
-
1
]
+
sum
[j
+
1
][i
]);
}
}
}
cout
<<
f
[main_len
-
1
][k
]
<< endl
;
return
0
;
}